3.775 \(\int \sec ^2(c+d x) (a+b \sec (c+d x))^2 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=198 \[ \frac{\left (4 a^2 B+6 a b C+3 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (4 a^2 B+6 a b C+3 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{\left (5 a (a C+2 b B)+4 b^2 C\right ) \tan ^3(c+d x)}{15 d}+\frac{\left (5 a (a C+2 b B)+4 b^2 C\right ) \tan (c+d x)}{5 d}+\frac{b (6 a C+5 b B) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac{b C \tan (c+d x) \sec ^3(c+d x) (a+b \sec (c+d x))}{5 d} \]

[Out]

((4*a^2*B + 3*b^2*B + 6*a*b*C)*ArcTanh[Sin[c + d*x]])/(8*d) + ((4*b^2*C + 5*a*(2*b*B + a*C))*Tan[c + d*x])/(5*
d) + ((4*a^2*B + 3*b^2*B + 6*a*b*C)*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (b*(5*b*B + 6*a*C)*Sec[c + d*x]^3*Tan[c
+ d*x])/(20*d) + (b*C*Sec[c + d*x]^3*(a + b*Sec[c + d*x])*Tan[c + d*x])/(5*d) + ((4*b^2*C + 5*a*(2*b*B + a*C)
)*Tan[c + d*x]^3)/(15*d)

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Rubi [A]  time = 0.351783, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {4072, 4026, 4047, 3767, 4046, 3768, 3770} \[ \frac{\left (4 a^2 B+6 a b C+3 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (4 a^2 B+6 a b C+3 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{\left (5 a (a C+2 b B)+4 b^2 C\right ) \tan ^3(c+d x)}{15 d}+\frac{\left (5 a (a C+2 b B)+4 b^2 C\right ) \tan (c+d x)}{5 d}+\frac{b (6 a C+5 b B) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac{b C \tan (c+d x) \sec ^3(c+d x) (a+b \sec (c+d x))}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + b*Sec[c + d*x])^2*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

((4*a^2*B + 3*b^2*B + 6*a*b*C)*ArcTanh[Sin[c + d*x]])/(8*d) + ((4*b^2*C + 5*a*(2*b*B + a*C))*Tan[c + d*x])/(5*
d) + ((4*a^2*B + 3*b^2*B + 6*a*b*C)*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (b*(5*b*B + 6*a*C)*Sec[c + d*x]^3*Tan[c
+ d*x])/(20*d) + (b*C*Sec[c + d*x]^3*(a + b*Sec[c + d*x])*Tan[c + d*x])/(5*d) + ((4*b^2*C + 5*a*(2*b*B + a*C)
)*Tan[c + d*x]^3)/(15*d)

Rule 4072

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 4026

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*(m + n
)), x] + Dist[1/(m + n), Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*Simp[a^2*A*(m + n) + a*b*B*n + (a
*(2*A*b + a*B)*(m + n) + b^2*B*(m + n - 1))*Csc[e + f*x] + b*(A*b*(m + n) + a*B*(2*m + n - 1))*Csc[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] &&  !
(IGtQ[n, 1] &&  !IntegerQ[m])

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^3(c+d x) (a+b \sec (c+d x))^2 (B+C \sec (c+d x)) \, dx\\ &=\frac{b C \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac{1}{5} \int \sec ^3(c+d x) \left (a (5 a B+3 b C)+\left (4 b^2 C+5 a (2 b B+a C)\right ) \sec (c+d x)+b (5 b B+6 a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{b C \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac{1}{5} \int \sec ^3(c+d x) \left (a (5 a B+3 b C)+b (5 b B+6 a C) \sec ^2(c+d x)\right ) \, dx+\frac{1}{5} \left (4 b^2 C+5 a (2 b B+a C)\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac{b (5 b B+6 a C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{b C \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac{1}{4} \left (4 a^2 B+3 b^2 B+6 a b C\right ) \int \sec ^3(c+d x) \, dx-\frac{\left (4 b^2 C+5 a (2 b B+a C)\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac{\left (4 b^2 C+5 a (2 b B+a C)\right ) \tan (c+d x)}{5 d}+\frac{\left (4 a^2 B+3 b^2 B+6 a b C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{b (5 b B+6 a C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{b C \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac{\left (4 b^2 C+5 a (2 b B+a C)\right ) \tan ^3(c+d x)}{15 d}+\frac{1}{8} \left (4 a^2 B+3 b^2 B+6 a b C\right ) \int \sec (c+d x) \, dx\\ &=\frac{\left (4 a^2 B+3 b^2 B+6 a b C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (4 b^2 C+5 a (2 b B+a C)\right ) \tan (c+d x)}{5 d}+\frac{\left (4 a^2 B+3 b^2 B+6 a b C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{b (5 b B+6 a C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{b C \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac{\left (4 b^2 C+5 a (2 b B+a C)\right ) \tan ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 1.51463, size = 150, normalized size = 0.76 \[ \frac{15 \left (4 a^2 B+6 a b C+3 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 \left (5 \left (a^2 C+2 a b B+2 b^2 C\right ) \tan ^2(c+d x)+15 \left (a^2 C+2 a b B+b^2 C\right )+3 b^2 C \tan ^4(c+d x)\right )+15 \left (4 a^2 B+6 a b C+3 b^2 B\right ) \sec (c+d x)+30 b (2 a C+b B) \sec ^3(c+d x)\right )}{120 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a + b*Sec[c + d*x])^2*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(15*(4*a^2*B + 3*b^2*B + 6*a*b*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*(4*a^2*B + 3*b^2*B + 6*a*b*C)*Sec[c
+ d*x] + 30*b*(b*B + 2*a*C)*Sec[c + d*x]^3 + 8*(15*(2*a*b*B + a^2*C + b^2*C) + 5*(2*a*b*B + a^2*C + 2*b^2*C)*
Tan[c + d*x]^2 + 3*b^2*C*Tan[c + d*x]^4)))/(120*d)

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Maple [A]  time = 0.046, size = 312, normalized size = 1.6 \begin{align*}{\frac{B{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{B{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,{a}^{2}C\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{4\,Bab\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,Bab\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{abC\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{2\,d}}+{\frac{3\,abC\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,abC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{B{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,{b}^{2}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,B{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{8\,{b}^{2}C\tan \left ( dx+c \right ) }{15\,d}}+{\frac{{b}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,{b}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+b*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

1/2/d*B*a^2*sec(d*x+c)*tan(d*x+c)+1/2/d*B*a^2*ln(sec(d*x+c)+tan(d*x+c))+2/3/d*a^2*C*tan(d*x+c)+1/3/d*a^2*C*tan
(d*x+c)*sec(d*x+c)^2+4/3/d*B*a*b*tan(d*x+c)+2/3/d*B*a*b*tan(d*x+c)*sec(d*x+c)^2+1/2/d*a*b*C*tan(d*x+c)*sec(d*x
+c)^3+3/4*a*b*C*sec(d*x+c)*tan(d*x+c)/d+3/4/d*a*b*C*ln(sec(d*x+c)+tan(d*x+c))+1/4/d*B*b^2*tan(d*x+c)*sec(d*x+c
)^3+3/8/d*B*b^2*sec(d*x+c)*tan(d*x+c)+3/8/d*B*b^2*ln(sec(d*x+c)+tan(d*x+c))+8/15*b^2*C*tan(d*x+c)/d+1/5/d*b^2*
C*tan(d*x+c)*sec(d*x+c)^4+4/15/d*b^2*C*tan(d*x+c)*sec(d*x+c)^2

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Maxima [A]  time = 1.01124, size = 373, normalized size = 1.88 \begin{align*} \frac{80 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} + 160 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b + 16 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C b^{2} - 30 \, C a b{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, B b^{2}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, B a^{2}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/240*(80*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 + 160*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a*b + 16*(3*tan(d*
x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*b^2 - 30*C*a*b*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d
*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 15*B*b^2*(2*(3*sin(d*
x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x +
c) - 1)) - 60*B*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d

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Fricas [A]  time = 0.544358, size = 521, normalized size = 2.63 \begin{align*} \frac{15 \,{\left (4 \, B a^{2} + 6 \, C a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (4 \, B a^{2} + 6 \, C a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (5 \, C a^{2} + 10 \, B a b + 4 \, C b^{2}\right )} \cos \left (d x + c\right )^{4} + 15 \,{\left (4 \, B a^{2} + 6 \, C a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} + 24 \, C b^{2} + 8 \,{\left (5 \, C a^{2} + 10 \, B a b + 4 \, C b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \,{\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/240*(15*(4*B*a^2 + 6*C*a*b + 3*B*b^2)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(4*B*a^2 + 6*C*a*b + 3*B*b^2
)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(16*(5*C*a^2 + 10*B*a*b + 4*C*b^2)*cos(d*x + c)^4 + 15*(4*B*a^2 +
6*C*a*b + 3*B*b^2)*cos(d*x + c)^3 + 24*C*b^2 + 8*(5*C*a^2 + 10*B*a*b + 4*C*b^2)*cos(d*x + c)^2 + 30*(2*C*a*b +
B*b^2)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (B + C \sec{\left (c + d x \right )}\right ) \left (a + b \sec{\left (c + d x \right )}\right )^{2} \sec ^{3}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(a+b*sec(d*x+c))**2*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Integral((B + C*sec(c + d*x))*(a + b*sec(c + d*x))**2*sec(c + d*x)**3, x)

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Giac [B]  time = 1.53993, size = 713, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/120*(15*(4*B*a^2 + 6*C*a*b + 3*B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*B*a^2 + 6*C*a*b + 3*B*b^2)*
log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(60*B*a^2*tan(1/2*d*x + 1/2*c)^9 - 120*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 2
40*B*a*b*tan(1/2*d*x + 1/2*c)^9 + 150*C*a*b*tan(1/2*d*x + 1/2*c)^9 + 75*B*b^2*tan(1/2*d*x + 1/2*c)^9 - 120*C*b
^2*tan(1/2*d*x + 1/2*c)^9 - 120*B*a^2*tan(1/2*d*x + 1/2*c)^7 + 320*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 640*B*a*b*ta
n(1/2*d*x + 1/2*c)^7 - 60*C*a*b*tan(1/2*d*x + 1/2*c)^7 - 30*B*b^2*tan(1/2*d*x + 1/2*c)^7 + 160*C*b^2*tan(1/2*d
*x + 1/2*c)^7 - 400*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 800*B*a*b*tan(1/2*d*x + 1/2*c)^5 - 464*C*b^2*tan(1/2*d*x +
1/2*c)^5 + 120*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 320*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 640*B*a*b*tan(1/2*d*x + 1/2*c
)^3 + 60*C*a*b*tan(1/2*d*x + 1/2*c)^3 + 30*B*b^2*tan(1/2*d*x + 1/2*c)^3 + 160*C*b^2*tan(1/2*d*x + 1/2*c)^3 - 6
0*B*a^2*tan(1/2*d*x + 1/2*c) - 120*C*a^2*tan(1/2*d*x + 1/2*c) - 240*B*a*b*tan(1/2*d*x + 1/2*c) - 150*C*a*b*tan
(1/2*d*x + 1/2*c) - 75*B*b^2*tan(1/2*d*x + 1/2*c) - 120*C*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 -
1)^5)/d